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Lecture 13
Analysis of Statically Indeterminate Structures by the Force Method- Beam- Frame- Truss- Composite Structures
Analysis of Statically Indeterminate Structures
by the Force Method
Lecture 13
In this chapter we will apply the force or flexibility method
(method of consistent deformations) to analyze statically
indeterminate trusses, beams, and frames.
The method, which was introduced by James C. Maxwell in 1864, essentially involves removing enough restraints from the indeterminate structure to render it statically determinate. This determinate structure, which must be statically stable, is referred to as the primary structure.
Statically Indeterminate Structures
Advantages & Disadvantages
For a given loading, the max stress and deflection of an indeterminate structure
are generally smaller than those of its statically determinate counterpart.
Statically indeterminate structure has a tendency to redistribute its load to its
redundant supports in cases of faulty designs or overloading.
Lecture 13
Although statically indeterminate structure can support loading with
thinner members & with increased stability compared to their
statically determinate counterpart, the cost savings in material must
be compared with the added cost to fabricate the structure since
often it becomes more costly to construct the supports & joints of
an indeterminate structure.
One has to careful of differential displacement of the supports as well.
Lecture 13
Lecture 13
Method of AnalysisTo satisfy equilibrium, compatibility and force-displacement requirements for the
structure.
1. Force Method
2. Displacement Method
Lecture 13
Force vs Displacement Methods• Force methods
• Find degree of Statically indeterminacy.
• Choose redundant forces
• Use compatibility conditions or least work principle to solve these redundant forces
• Displacement methods• Choose degrees of freedom (DOFs: displacement or rotation angles)
• Relate internal forces to DOFs
• Use equilibrium to solve DOFs
• Obtain internal forces from DOFs
Lecture 13
By
qB
Force Method of Analysis: General Procedure
Consider the beam shown in Fig.
From free-body diagram, there would be 4 unknown support
reactions 3 equilibrium equations. The Beam is
indeterminate to first degree to obtain the additional
equation, use principle of superposition & consider the
compatibility of displacement at one of the supports.
This is done by choosing one of the support reactions as
redundant & temporarily removing its effect on the beam
Lecture 13
Lecture 13
This will allow the beam to be
statically determinate and stable.
Here, we will remove the rocker at B.
As a result, the load P will cause B to
be displaced downward.
By superposition, the unknown
reaction at B causes the beam at B
to be displaced upward.
Assuming positive displacements act upward, we write the necessary compatibility equation at the rocker as:
Lecture 13
Δ'BB - Upward displacement at BBy - Unknown reaction at BfBB - Linear flexibility coefficient
Substitution Eq. (2) into Eq.(1), we get:
Using methods in Chapter 8 or 9 to solve for ΔB and fBB, Bycan be found.
Reactions at wall A can then be determined from equation of equilibrium.
The choice of redundant is arbitrary.
Lecture 13
The moment at A, Fig. can be determined directly by removing the capacity of the beam to support moment at A, replacing fixed support by pin support.
As shown in Fig., the rotation at A caused by P is θA.
The rotation at A caused by the redundant MA at A is θ’AA.
Lecture 13
AAAA
AAAAA
M
M
αθ0
:requiresity Compatibil
α'θ Similarly,
+=
=
In this case, MA = -θA/αAA, a negative value,
which simply means that acts in the opposite
direction to the unit couple moment.
Lecture 13
the beam is indeterminate to the second degree andtherefore two compatibility equations will be necessary for the solution. We will choose the vertical forces at the roller supports, B and C, as redundants. The resultant statically determinate beam deflects as shown when the redundants are removed. Each redundant force, which is assumed to act downward, deflects this beam as shown in Fig.
Lecture 13
CCyCByC
BCyBByB
fCfB
fCfB
++=+
++=+
Δ0↓
Δ0↓
Once the load-displacement relations are established using the
methods of Chapter 8 or 9, these equations may be solved
simultaneously for the two unknown forces By and Cy
Lecture 13
PROCEDURE FOR ANALYSIS
• Principle of Superposition. Determine the number of degrees n to which the structure is indeterminate. Then specify the n unknown redundant forces or moments that must be removed from the structure in order to make it statically determinate and stable. Using the principle of superposition, draw the statically indeterminate structure and show it to be equal to a series of corresponding statically determinate structures. The primary structure supports the same external loads as the statically indeterminate structure, and each of the other structures added to the primary structure shows the structure loaded with a separate redundant force or moment. Also, sketch the elastic curve on each structure and indicate symbolically the displacement or rotation at the point of each redundant force or moment.
Lecture 13
• Compatibility Equations. Write a compatibility equation for the displacement or rotation at each point where there is a redundant force or moment. These equations should be expressed in terms of the unknown redundants and their corresponding flexibility coefficients obtained from unit loads or unit couple moments that are collinear with the redundant forces or moments.
Determine all the deflections and flexibility coefficients using the table on the inside front cover or the methods of Chapter 8 or 9. Substitute these load-displacement relations into the compatibility equations and solve for the unknown redundants. In particular, if a numerical value for a redundant is negative, it indicates the redundant acts opposite to its corresponding unit force or unit couple moment.
• Equilibrium Equations. Draw a free-body diagram of the structure. Since the redundant forces and/or moments have been calculated, the remaining unknown reactions can be determined from the equations of equilibrium.
Lecture 13
Maxwell’s Theorem of Reciprocal Displacement: Betti’s LawThe displacement of a point B on a structure due to a unit
load acting at point A is equal to the displacement of point A when the load is acting at point B.
Proof of this theorem is easily demonstrated using the principle of virtual work.
Lecture 13
ABBA ff =
The theorem also applies for reciprocal rotations.
The rotation at point B on a structure due to a unit couple
moment acting at point A is equal to the rotation at
point A when the unit couple is acting at point B.
∫ dxEI
mmf AB
BA =
∫ dxEI
mmf BA
AB =
When a real unit load acts at A, assume that the internal moments
in the beam are represented by mA. To determine the flexibility
coefficient at B, that is, fBA , a virtual unit load is placed at B, and the
internal moments mB are computed.
Likewise, if the flexibility coefficient fAB is to be determined when a real unit load acts at B, then mB
represents the internal moments in the beam due to a real unit load. Furthermore, mA represents the internal moments due to a virtual unit load at A.
Lecture 13
Force Method of Analysis: Beams
Lecture 13
Force Method of Analysis: Frames
Lecture 13
Force Method of Analysis: Trusses
The degree of indeterminacy of a truss can usually be determined
by inspection; however, if this becomes difficult b + r> 2j. Here the
unknowns are represented by the number of bar forces (b) plus
the support reactions (r), and the number of available equilibrium
equations is 2j since two equations can be written for each of the
(j) joints.
The force method is quite suitable for analyzing trusses that are
statically indeterminate to the first or second degree.
Lecture 13
Lecture 13
Composite Structures
Composite structures are composed of some members
subjected only to axial force, while other members are
subjected to bending. If the structure is statically
indeterminate, the force method can conveniently be
used for its analysis. The following example illustrates
the procedure.
Lecture 13
Additional Remarks on the Force Method of Analysis
Now that the basic ideas regarding the force method have been
developed, we will proceed to generalize its application and
discuss its usefulness.
When computing the flexibility coefficients, fij (or aij), for the
structure, it will be noticed that they depend only on the material
and geometrical properties of the members and not on the
loading of the primary structure. Hence these values, once
determined, can be used to compute the reactions for any
loading.
For a structure having n redundant reactions, Rn, we can write n
compatibility equations, namely:Lecture 13
Here the displacements, Δ1 ,..., Δn, are caused by both the real
loads on the primary structure and by support settlement or
dimensional changes due to temperature differences or
fabrication errors in the members.
Lecture 13
0Δ
02Δ
0Δ
2211
2222121
12121111
=++++
=++++
=++++
nnnnnn
nn
nn
RfRfRf
RfRfRf
RfRfRf
To simplify computation for structures having a large degree of indeterminacy, the above equations can be recast into a matrix form,
Lecture 13
In particular, note that fij = fji(f12= f21 , etc.), a consequence of
Maxwell's theorem of reciprocal displacements (or Betti's law).
Hence the flexibility matrix will be symmetric, and this feature is
beneficial when solving large sets of linear equations, as in the case
of a highly indeterminate structure.
Symmetric StructuresA structural analysis of any highly indeterminate structure, or for that
matter, even a statically determinate structure, can be simplified
provided the designer or analyst can recognize those structures that
are symmetric and support either symmetric or antisymmetric
loadings. In a general sense, a structure can be classified as being
symmetric provided half of it develops the same internal loadings and
deflections as its mirror image reflected about its central axis.
Normally symmetry requires the material composition, geometry,
supports, and loading to be the same on each side of the structure.
However, this does not always have to be the case. Notice that for
horizontal stability a pin is required to support the beam and truss in
Figs. Here the horizontal reaction at the pin is zero, and so both of
these structures will deflect and produce the same internal loading as
their reflected counterpart. As a result, they can be classified as being
symmetric. Lecture 13
Lecture 13
Realize that this would not be the case for the frame, if the fixed support at A was replaced by a pin, since then the deflected shape and internal loadings would not be the same on its left and right sides.
Sometimes a symmetric structure supports an antisymmetric loading, that is, the loading on its reflected side has the opposite direction, such as shown by the two examples in Fig. 10-19. Provided the structure is symmetric and its loading is either symmetric or antisymmetric, then a structural analysis will only have to be performed on half the members of the structure since the same (symmetric) or opposite (antisymmetric) results will be produced on the other half. If a structure is symmetric and its applied loading is unsymmetrical, then it is possible to transform this loading into symmetric and antisymmetric components. To do this, the loading is first divided in half, then it is reflected to the other side of the structure and both symmetric and antisymmetric components are produced. For example, the loading on the beam in Fig. 10-20a is divided by two and reflected about the beam's axis of symmetry. From this, the symmetric and antisymmetric components of the load are produced as shown in Fig. 10-206. When added together these components produce the original loading. A separate structural analysis can now be performed using the symmetric and antisymmetric loading components and the results superimposed to obtain the actual behavior of the structure.
Lecture 13
What Have You Learnt?
•Degree of statically indeterminacy.
•Compatibility equations.
•Analysis of the beam, frame, and truss by using force method.
Lecture 13
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